1988 AHSME Problems/Problem 28

Problem

An unfair coin has probability $p$ of coming up heads on a single toss. Let $w$ be the probability that, in $5$ independent toss of this coin, heads come up exactly $3$ times. If $w = 144 / 625$ , then

$\textbf{(A)}\ p\text{ must be }\tfrac{2}{5}\qquad \textbf{(B)}\ p\text{ must be }\tfrac{3}{5}\qquad\\ \textbf{(C)}\ p\text{ must be greater than }\tfrac{3}{5}\qquad \textbf{(D)}\ p\text{ is not uniquely determined}\qquad\\ \textbf{(E)}\ \text{there is no value of } p \text{ for which }w =\tfrac{144}{625}$

Solution

We have $w = {5\choose3}p^{3}(1-p)^{2} = 10p^{3}(1-p)^{2}$ , so we need to solve $10p^{3}(1-p)^{2} = \frac{144}{625} \implies p^{3}(1-p)^{2} = \frac{72}{3125}$ . Now observe that when $p=0$ , the left-hand side evaluates to $0$ , which is less than $\frac{72}{3125}$ ; when $p=\frac{1}{2}$ , it evaluates to $\frac{1}{32}$ , which is more than $\frac{72}{3125}$ , and when $p=1$ , it evaluates to $0$ again. Thus, since $p^{3}(1-p)^{2}$ is continuous, the Intermediate Value Theorem tells us there is a solution between $0$ and $\frac{1}{2}$ and another solution between $\frac{1}{2}$ and $1$ , meaning that there is not a unique value of $p$ , so the answer is $\boxed{\text{D}}$ .

1988 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1988 AHSME Problems/Problem 28

Problem

Solution

See also